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+ | = [[ECE_PhD_Qualifying_Exams|ECE Ph.D. Qualifying Exam]] in Communication Networks Signal and Image processing (CS), Question 5, August 2011= | ||
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+ | ==Question== | ||
+ | '''Part 1. ''' 50 pts | ||
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− | + | <font color="#ff0000"><span style="font-size: 19px;"><math>\color{blue}\text{Consider the following discrete space system with input } x(m,n) \text{ and output } y(m,n). | |
+ | </math></span></font> | ||
− | + | <math>\color{blue} | |
+ | y(m,n) = \sum_{k=-\infty}^{\infty}{\sum_{l=-\infty}^{\infty}{x(m-k,n-l)h(k,l)}}. | ||
+ | </math><br> | ||
− | <math> | + | <math>\color{blue} |
+ | \text{For parts a) and b) let} | ||
+ | </math><br> | ||
+ | <math>\color{blue} | ||
+ | h(m,n)=sinc(mT,nT), \text{where} T\leq1. | ||
+ | </math><br> | ||
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− | <math>\ | + | <math>\color{blue}\text{a) Calculate the frequency response, }H \left( e^{j\mu},e^{j\nu} \right).</math><br> |
− | <math> | + | <math>\color{blue}\text{b) Sketch the frequency response for } |\mu| < 2\pi \text{ and } |nu| < 2\pi \text{ when } T = \frac{1}{2} |
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</math><br> | </math><br> | ||
− | <math>\color{blue}\text{ | + | <math>\color{blue} |
+ | \text{For parts c), d), and e) let} | ||
+ | </math><br> | ||
+ | <math>\color{blue} | ||
+ | h(m,n)=sinc\left( \frac{(n+m)T}{\sqrt[]{2}},\frac{(n-m)T}{\sqrt[]{2}} \right) | ||
+ | </math><br> | ||
+ | <math>\color{blue} | ||
+ | \text{where } T\leq1. | ||
+ | </math><br> | ||
− | + | <math>\color{blue}\text{c) Sketch the frequency response for } |\mu| < 2\pi \text{ and } |nu| < 2\pi \text{ when } T = \frac{1}{2} | |
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</math><br> | </math><br> | ||
− | + | <math>\color{blue}\left( \text{d} \right) \text{Find } | |
− | + | f_{Y}\left(y|z \right ). | |
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− | <math>\color{blue}\left( \text{d} \right) \text{Find} | + | |
− | f_{Y}\left(y|z \right ) | + | |
</math><br> | </math><br> | ||
− | + | <math>\color{blue}\left( \text{e} \right) \text{Find } | |
− | + | f_{XY}\left(x,y|z \right ). | |
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− | <math>\color{blue}\left( \text{e} \right) \text{Find} | + | |
− | f_{XY}\left(x,y|z \right ) | + | |
</math><br> | </math><br> | ||
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+ | :'''Click [[ECE-QE_CS1-2011_solusion-1|here]] to view student [[ECE-QE_CS1-2011_solusion-1|answers and discussions]]''' | ||
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+ | '''Part 2.''' 25 pts | ||
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− | + | <font color="#ff0000"><span style="font-size: 19px;"><math>\color{blue} \text{Show that if a continuous-time Gaussian random process } \mathbf{X}(t) \text{ is wide-sense stationary, it is also strict-sense stationary.} | |
− | ---- | + | </math></span></font> |
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+ | :'''Click [[ECE-QE_CS1-2011_solusion-2|here]] to view student [[ECE-QE_CS1-2011_solusion-2|answers and discussions]]''' | ||
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− | + | [[ECE_PhD_Qualifying_Exams|Back to ECE Qualifying Exams (QE) page]] | |
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Revision as of 11:20, 27 July 2012
ECE Ph.D. Qualifying Exam in Communication Networks Signal and Image processing (CS), Question 5, August 2011
Question
Part 1. 50 pts
$ \color{blue}\text{Consider the following discrete space system with input } x(m,n) \text{ and output } y(m,n). $
$ \color{blue} y(m,n) = \sum_{k=-\infty}^{\infty}{\sum_{l=-\infty}^{\infty}{x(m-k,n-l)h(k,l)}}. $
$ \color{blue} \text{For parts a) and b) let} $
$ \color{blue} h(m,n)=sinc(mT,nT), \text{where} T\leq1. $
$ \color{blue}\text{a) Calculate the frequency response, }H \left( e^{j\mu},e^{j\nu} \right). $
$ \color{blue}\text{b) Sketch the frequency response for } |\mu| < 2\pi \text{ and } |nu| < 2\pi \text{ when } T = \frac{1}{2} $
$ \color{blue} \text{For parts c), d), and e) let} $
$ \color{blue} h(m,n)=sinc\left( \frac{(n+m)T}{\sqrt[]{2}},\frac{(n-m)T}{\sqrt[]{2}} \right) $
$ \color{blue} \text{where } T\leq1. $
$ \color{blue}\text{c) Sketch the frequency response for } |\mu| < 2\pi \text{ and } |nu| < 2\pi \text{ when } T = \frac{1}{2} $
$ \color{blue}\left( \text{d} \right) \text{Find } f_{Y}\left(y|z \right ). $
$ \color{blue}\left( \text{e} \right) \text{Find } f_{XY}\left(x,y|z \right ). $
- Click here to view student answers and discussions
Part 2. 25 pts
$ \color{blue} \text{Show that if a continuous-time Gaussian random process } \mathbf{X}(t) \text{ is wide-sense stationary, it is also strict-sense stationary.} $
- Click here to view student answers and discussions